Which of the following cannot act as a ligand:
In the complexes [Fe(CN)6]3– and [Co(en)3]3+, the coordination number of iron and cobalt are respectively (en = ethylenediamine):
In 1st compound CN is an unidentate ligand and hence it will surround Fe atom from 6 different sides
On the other hand, 2nd compound contains bidentate ligand so each (en) will have two different site of donating electrons and hence it required 3 (en) atoms for surrounding Co atom from 6 different sides.
Hence A is correct.
The oxidation number, coordination number and magnetic moment in the following complex is:
x + (-2 x 2) + 0 x 2 = -1
x = +3
Coordination number is 6, because C2O42- bidentate ligand and NH3 monodentate.
Magnetic moment = √n(n+2)
where n is no. of unpaired electrons.
For Cr , n = 3 so Magnetic moment will be √15 BM
As a ligand Cl– is:
Cl- is π-donor (minus charge form σ bond with the metal orbitals, eg, t1u and a1g orbitals in Oh field. The lone pair of electron on Cl will try to form π-bonding with t2g orbitals of metal, electron cloud is transferred from Cl to metal d orbitals).
The CFSE for the following d3 metal ions (V2+, Cr3+, Mo3+) decrease in the following order :
Octahedral complex of Cr (III) will be:
Which of the following is a spin paired complex ion?
Correct Answer :- d
Explanation : In [Co(NH3)6]3+, the oxidation state of cobalt is +3.Ammonia is a strong field ligand so it pair up 4 unpaired electron and free up 2 3−d orbitals. These 3−d orbitals are involved in hybridisation with one 4S and three 4P orbitals forming an inner orbital complex, so hybridisation of [Co(NH3)6]3+ is d2sp3
Compound ‘X’ has molecular formula CrCl2Br.6H2O can show type of isomerism:
(I) Hydrate iso merism.
(II) Ionizat ion isomerism.
(III) Geometrical isomerism.
(IV) Optical isomerism.
What is the change in oxidation state of cobalt in the following reaction?
[Co(NH3)4 Cl2]+ + H2O → [Co(NH3)4 (H2O)Cl]2+ + Cl–
The first row transition metal complexes having tetrahedral geometry are high–spin due to:
The zero magnetic moment of octahedral K2NiF6 is due to:
⇒x−6 = −2
x = + 4
In +4 oxidation state, fluoride also behaves as a strong field ligand
⇒ low spin d6 Ni(IV) complex
⇒ unpaired electrons = 0
⇒ zero magnetic moment
What is the spin only magnetic moment value in (Bohr Magneton units) of Cr(CO)6?
The electron configuration is [Ar]3d54s1.We have to accomodate the 6 Ligands and the fact that CO is a strong ligand.
This results in d2sp3 hybridization. Therefore, there are no unpaired electrons in Cr(CO)6. Hence n=0
And the spin only magnetic moment is also 0.
The volume (in mL) of 0.1 M AgNO3 required for complex precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to
Which statement(s) is (are) true:
(I) High spin complexes are always paramagnetic
(II) Low spin complexes are always diamagnetic
(III) H2O is more likely to form a low spin complex than CN–
(IV) Tetrahedral complexes are more likely to be low spin than high spin
Correct Answer :- C
Explanation : a) High-spin complexes usually contain more unpaired electrons because the pairing energy is larger than the splitting energy. With more unpaired electrons, high-spin complexes are often paramagnetic.
b) Low-spin complexes contain more paired electrons because the splitting energy is larger than the pairing energy. These complexes, such as [Fe(CN)6]3-, are more often diamagnetic or weakly paramagnetic.
When EDTA solution is added to Mg2+ ion solution, then which of the following statements is not true?
Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)]is
Which one the following octahedral complexes will be distorted:
Because of more number i.e. 4 e- (e- e- repulsion) which makes it more distorted. But in case of Mn & Fe both of them has 3d5 configuration which is more stable & decreases the chances of distortion. In case of Cr3+ less no of e- are present, so, less e- e- repulsion hence lesser will be the distortion.
When there is unsymmetrical filling of electron in Eg orbital occur the shape will distorted e.g. in the case of d4. This is known as Jahn-Teller distortion.
Hence A is correct.
The crystal field stabilization energy (CFSE), will be the highest for:
The complex with maximum CFSE is:
The compound which exhibits Jahn-Teller distortion is: